Let A and b are two square idempotent matrices such that `ABpm BA` is a null matrix, the value of det (A - B)
cann vbe equal

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \(\det(A - B)\) given that \(A\) and \(B\) are idempotent matrices and that \(AB + BA = 0\).

### Step-by-Step Solution:

1. **Understanding Idempotent Matrices**:
- A matrix \(A\) is idempotent if \(A^2 = A\).
- Similarly, for matrix \(B\), we have \(B^2 = B\).

2. **Using the Given Condition**:
- We know that \(AB + BA = 0\). This implies that \(AB = -BA\).

3. **Finding \(\det(A - B)\)**:
- We can express \(\det(A - B)\) using the property of determinants:
\[ \det(A - B) = \det(A + (-B)) \]

4. **Using the Idempotent Property**:
- Since \(A\) and \(B\) are idempotent, we can write:
\[ A^2 = A \quad \text{and} \quad B^2 = B \]
- Therefore, we can expand \(\det(A - B)\) as follows:
\[ \det(A - B) = \det(A^2 - AB - BA + B^2) \]

5. **Substituting the Idempotent Properties**:
- Substitute \(A^2 = A\) and \(B^2 = B\):
\[ \det(A - B) = \det(A - AB - BA + B) \]
- Since \(AB + BA = 0\), we have:
\[ \det(A - B) = \det(A + B) \]

6. **Finding the Determinant**:
- Now we can find \(\det(A + B)\). Since \(AB + BA = 0\), we can conclude that:
\[ \det(A + B) = 0 \quad \text{or} \quad 1 \]
- This leads us to:
\[ \det(A - B)^2 = \det(A + B) \]

7. **Conclusion**:
- Therefore, we have:
\[ \det(A - B) = 0, 1, \text{ or } -1 \]

### Final Result:
The value of \(\det(A - B)\) can be equal to \(0\), \(1\), or \(-1\).